Yes.

Let $\nabla$ be an arbitrary connection on the tangent bundle of a Riemannian manifold $(M,g)$. The standard trick for expressing the Levi-Civita connection in terms of $g$ gives you, for any 3 vector fields $X$, $Y$, $Z$: $$Xg(Y,Z)+ Yg(Z,X)- Zg(X,Y)= N(X,Y,Z) $$ $$+ g(T(X,Z),Y)+ g(T(Y,Z),X)- g(T(X,Y),Z) $$ $$ +2 g(\nabla_X Y,Z)- g([X,Y],Z) + g([X,Z],Y) + g([Y,Z],X),$$

where $$ N(X,Y,Z)= \nabla_Xg(Y,Z)+ \nabla_Yg(Z,X)-\nabla_Zg(X,Y). $$ This is the "non-metricity": $N=0\Leftrightarrow \nabla g=0$.

Now, turning to the case at hand: we define the $\pm$ and $0$ connections by $$ (\nabla_X Y)_e=\epsilon [X,Y],$$ $ \epsilon = 1, 0, \frac{1}{2}$ respectively, so the torsion is $$T(X,Y) = (2\epsilon -1)[X,Y]= \pm[X,Y]\textrm{ or } 0, $$ hence the names of the connections. But then you get $$ 0 = N(X,Y,Z) -2\epsilon\left[ g([Z,Y],X) + g(Y,[Z,X]) \right],$$ and the second summand is zero due to bi-invariance, so $N=0$.

Yes.

Let $\nabla$ be an arbitrary connection on the tangent bundle of a Riemannian manifold $(M,g)$. The standard trick for expressing the Levi-Civita connection in terms of $g$ gives you, for any 3 vector fields $X$, $Y$, $Z$: $$Xg(Y,Z)+ Yg(Z,X)- Zg(X,Y)= N(X,Y,Z) $$ $$+ g(T(X,Z),Y)+ g(T(Y,Z),X)- g(T(X,Y),Z) $$ $$ +2 g(\nabla_X Y,Z)- g([X,Y],Z) + g([X,Z],Y) + g([Y,Z],X),$$

where $$ T(X,Y)=\nabla_X Y- \nabla_Y X -[X,Y]$$ is the torsion of $\nabla$ and $$ N(X,Y,Z)= \nabla_Xg(Y,Z)+ \nabla_Yg(Z,X)-\nabla_Zg(X,Y). $$ This is the "non-metricity": $N=0\Leftrightarrow \nabla g=0$.

Now, turning to the case at hand: we define the $\pm$ and $0$ connections by $$ (\nabla_X Y)_e=\epsilon [X,Y],$$ $ \epsilon = 1, 0, \frac{1}{2}$ respectively, so the torsion is $$T(X,Y) = (2\epsilon -1)[X,Y]= \pm[X,Y]\textrm{ or } 0, $$ hence the names of the connections. But then you get $$ 0 = N(X,Y,Z) -2\epsilon\left[ g([Z,Y],X) + g(Y,[Z,X]) \right],$$ and the second summand is zero due to bi-invariance, so $N=0$.

Yes.

Let $\nabla$ be an arbitrary connection on the tangent bundle of a Riemannian manifold $(M,g)$. The standard trick for expressing the Levi-Civita connection in terms of $g$ gives you, for any 3 vector fields $X$, $Y$, $Z$: $$Xg(Y,Z)+ Yg(Z,X)- Zg(X,Y)= N(X,Y,Z) + g(T(X,Z),Y)+ g(T(Y,Z),X)$$ $$- g(T(X,Y),Z) + 2g(\nabla_X Y,Z)- g([X,Y],Z) + g([X,Z],Y) + g([Y,Z],X),$$ where $$ N(X,Y,Z)= \nabla_Xg(Y,Z)+ \nabla_Yg(Z,X)-\nabla_Zg(X,Y). $$ This is the "non-metricity": $N=0\Leftrightarrow \nabla g=0$.

Now, turning to the case at hand: we define the $\pm$ and $0$ connections by $$ (\nabla_X Y)_e=\epsilon [X,Y],$$ $ \epsilon = 1, 0, \frac{1}{2}$ respectively, so the torsion is $$T(X,Y) = (2\epsilon -1)[X,Y]= \pm[X,Y]\textrm{ or } 0, $$ hence the names of the connections. But then you get $$ 0 = N(X,Y,Z) -2\epsilon\left[ g([Z,Y],X) + g(Y,[Z,X]) \right],$$ and the second summand is zero due to bi-invariance, so $N=0$.

Yes.

Let $\nabla$ be an arbitrary connection on the tangent bundle of a Riemannian manifold $(M,g)$. The standard trick for expressing the Levi-Civita connection in terms of $g$ gives you, for any 3 vector fields $X$, $Y$, $Z$: $$Xg(Y,Z)+ Yg(Z,X)- Zg(X,Y)= N(X,Y,Z) $$ $$+ g(T(X,Z),Y)+ g(T(Y,Z),X)- g(T(X,Y),Z) $$ $$ +2 g(\nabla_X Y,Z)- g([X,Y],Z) + g([X,Z],Y) + g([Y,Z],X),$$

where $$ N(X,Y,Z)= \nabla_Xg(Y,Z)+ \nabla_Yg(Z,X)-\nabla_Zg(X,Y). $$ This is the "non-metricity": $N=0\Leftrightarrow \nabla g=0$.

Now, turning to the case at hand: we define the $\pm$ and $0$ connections by $$ (\nabla_X Y)_e=\epsilon [X,Y],$$ $ \epsilon = 1, 0, \frac{1}{2}$ respectively, so the torsion is $$T(X,Y) = (2\epsilon -1)[X,Y]= \pm[X,Y]\textrm{ or } 0, $$ hence the names of the connections. But then you get $$ 0 = N(X,Y,Z) -2\epsilon\left[ g([Z,Y],X) + g(Y,[Z,X]) \right],$$ and the second summand is zero due to bi-invariance, so $N=0$.